Question

accepted_bids = Bid.objects.filter(shipment__user=u, status='acc').select_related('shipment')
completed_shipments = []
for b in accepted_bids:
    completed_shipments.append(b.shipment)
vehicles_shipped = []
for s in completed_shipments:
    vehicles_shipped.extend(s.items.all())

最后，我想要一份已发货的车辆清单。如果车辆已完成装运，则装运车辆。如果已接受出价，则已完成装运。

我不希望迭代查询集，从而在必要之前强制命中数据库......例如，是不是有办法从出价列表中获取所有关联的货件？

这是我模特的精简版：

class Shipment(Model):
    user = ForeignKey(User, related_name='shipments')

class ShipmentItem(models.Model):
    shipment = ForeignKey(Shipment, related_name='items')

    class Meta:
    abstract = True

class VehicleItem(ShipmentItem):
    pass

class Bid(Model):
    user = ForeignKey(User, related_name='bids')
    shipment = ForeignKey(Shipment, related_name='bids')

Answer 1

提供架构很重要。但这是一个猜测：

Vehicle.objects.filter(shipment__bid__status='acc', shipment__user=u)

有没有办法简化这个Django查询？

1 个答案: